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Observe the Following Pattern 1 2 = 1 6 [ 1 × ( 1 + 1 ) × ( 2 × 1 + 1 ) ] 1 2 + 2 2 = 1 6 [ 2 × ( 2 + 1 ) × ( 2 × 2 + 1 ) ] 1 2 + 2 2 + 3 2 = 1 6 [ 3 × ( 3 + 1 ) × ( 2 × 3 + 1 ) ] 1 - Mathematics

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Question

Observe the following pattern \[1^2 = \frac{1}{6}\left[ 1 \times \left( 1 + 1 \right) \times \left( 2 \times 1 + 1 \right) \right]\]
\[ 1^2 + 2^2 = \frac{1}{6}\left[ 2 \times \left( 2 + 1 \right) \times \left( 2 \times 2 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 = \frac{1}{6}\left[ 3 \times \left( 3 + 1 \right) \times \left( 2 \times 3 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 + 4^2 = \frac{1}{6}\left[ 4 \times \left( 4 + 1 \right) \times \left( 2 \times 4 + 1 \right) \right]\] and find the values :  

52 + 62 + 72 + 82 + 92 + 102 + 112 + 122

 

 

Solution

Observing the six numbers on the RHS of the equalities:
The first equality, whose biggest number on the LHS is 1, has 1, 1, 1, 2, 1 and 1 as the six numbers.
The second equality, whose biggest number on the LHS is 2, has 2, 2, 1, 2, 2 and 1 as the six numbers.
The third equality, whose biggest number on the LHS is 3, has 3, 3, 1, 2, 3 and 1 as the six numbers.
The fourth equality, whose biggest number on the LHS is 4, has numbers 4, 4, 1, 2, 4 and 1 as the six numbers.
Note that the fourth number on the RHS is always 2 and the sixth number is always 1. The remaining numbers are equal to the biggest number on the LHS.
Hence, if the biggest number on the LHS is n, the six numbers on the RHS would be nn, 1, 2, n and 1.
Using this property, we can calculate the sums for (i) and (ii) as follows: 

\[ 1^2 + 2^2 + . . . . . . . + {10}^2 = \frac{1}{6} \times \left[ 10 \times \left( 10 + 1 \right) \times \left( 2 \times 10 + 1 \right) \right]\]
\[= \frac{1}{6} \times \left[ 10 \times 11 \times 12 \right] \]
\[ = 385\]

 

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Chapter 3: Squares and Square Roots - Exercise 3.2 [Page 20]

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RD Sharma Mathematics [English] Class 8
Chapter 3 Squares and Square Roots
Exercise 3.2 | Q 10.2 | Page 20

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